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mojhsa [17]
3 years ago
6

Graph the following piecewise function and then find the range.

Mathematics
2 answers:
alex41 [277]3 years ago
8 0

Answer:

The correct option is 3.

Step-by-step explanation:

The given piecewise function is

f(x)=\begin{cases}3x^2+1 & \text{ if } -4

Range is the set of output or y values.

The given function for 6 ≤ x < 9 is

f(x)=6

It is a constant function, the value of function is 6 for all values of x.

Range = 6

The given function for -4 < x < 6 is

f(x)=3x^2+1          .... (1)

It is a quadratic function.

The vertex form of a quadratic function is

f(x)=a(x-h)^2+k         ....(2)

Where (h,k) is vertex and a is constant.

From (1) and (2), we get a=3,h=0,k=1.

The vertex of this function is (0,1), it means the range of this function is greater than or equal to 1. But this function is only defined for -4 < x < 6.

f(-4)=3(-4)^2+1=49

f(6)=3(6)^2+1=109

The maximum value of the maximum value of the function is 109 at x=6. Since 6 is not included in the interval -4 < x < 6, therefore 109 is not included in the range.

Range = [1,109)

When we combined the range of both functions we get

Range = [1,109)

Therefore the correct option is 3.

Orlov [11]3 years ago
3 0
The same function as in the one of other your questions.

D:[1,109)

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A company manufactures running shoes and basketball shoes. The total revenue (in thousands of dollars) from x1 units of running
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Answer:

x_1 =2 , x_2=7

Step-by-step explanation:

Consider the revenue function given by R(x_1,x_2) = -5x_1^2-8x_2^2 -2x_1x_2+34x_1+116x_2. We want to find the values of each of the variables such that the gradient( i.e the first partial derivatives of the function) is 0. Then, we have the following (the explicit calculations of both derivatives are omitted).

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\frac{d^2R}{dx_1dx_2}= -2 = \frac{d^2R}{dx_2dx_1}

\frac{d^2R}{dx_{1}^2}=-10, \frac{d^2R}{dx_{2}^2}=-16

We have the following matrix,  

\left[\begin{matrix} -10 & -2 \\ -2 & -16\end{matrix}\right].

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